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binarno stablo pomocu polja.h

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zadatak4

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  1. #include<iostream>
  2. using namespace std;
  3.  
  4. struct anode{
  5.     int label;
  6.     bool used;
  7. };
  8.  
  9. struct atree{
  10.     anode polje[1000];
  11. };
  12.  
  13. atree *aInitB(int x,atree *T){
  14.     T = new atree;
  15.     for(int i=0;i<1000;i++){
  16.             T->polje[i].used = false;
  17.             T->polje[i].label = -1;
  18.             }
  19.     T->polje[1].label = x;
  20.     T->polje[1].used = true;
  21.     return T;
  22.     }
  23.  
  24. int aLabelB(int n,atree *T){
  25.     return T->polje[n].label;
  26.     }
  27.  
  28. void aChangeLabelB(int x,int n,atree *T){
  29.     T->polje[n].label = x;
  30.     }
  31.  
  32. int aRootB(atree *T){
  33.     return T->polje[1].label;
  34.     }
  35.  
  36. int aLeftChildB(int n,atree *T){
  37.     return 2*n;
  38.     }
  39.  
  40. int aRightChildB(int n,atree *T){
  41.     return 2*n+1;
  42.     }
  43.  
  44. int aParentB(int n,atree *T){
  45.     if(T->polje[1].label==n) return -1;
  46.     if(n%2) return n/2+1;
  47.     else return n/2;
  48.     }
  49.  
  50. void aCreateLeftB(int x,int n,atree *T){
  51.     if(T->polje[2*n].used) cout << "Lijevo dijete ve����¯�¿�½ postoji!" << endl;
  52.     else{
  53.         T->polje[2*n].label = x;
  54.         T->polje[2*n].used = true;
  55.         }
  56.     }
  57.  
  58. void aCreateRightB(int x,int n,atree *T){
  59.     if(T->polje[2*n+1].used) cout << "Desno dijete ve����¯�¿�½ postoji!" << endl;
  60.     else{
  61.         T->polje[2*n+1].label = x;
  62.         T->polje[2*n+1].used = true;
  63.         }
  64.     }
  65.  
  66. void aDeleteB(int n,atree *T){
  67.     if(T->polje[aLeftChildB(n,T)].used) aDeleteB(2*n,T);
  68.     if(T->polje[aRightChildB(n,T)].used) aDeleteB(2*n-1,T);
  69.     T->polje[n].label = -1;
  70.     T->polje[n].used = false;
  71.     }

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